I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation: $$ f(z) = \frac{3z+2}{4z+3} $$ In particular, I'd like to compute the new center and radius.

The Möbius transformation can be turned into inversion as well:

  • $C_1= 4|z|^2+3\overline{z}-3z-2 $
  • $C_2 =|z|^2 - 1$

Or we could turn the second circle into a fractional lineartrasnforation $g(z) = - \frac{1}{z}$. Then I could multiply the two transformations: $$ \left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 2 & 3 \\ 3 & 4 \end{array} \right] $$

and this could turn back into a circle:

  • $ C_1C_2 = 3|z|^2 + 4 \overline{z} + 3z + 2 $

I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.

  • $\begingroup$ Why is accepted solution more usefull than mine? $\endgroup$ – Aqua Jan 12 at 9:58

I would take three points on $\mid z\mid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=\frac57\,,f(-1)=1$ and $f(i)=\frac{18+i}{25}$.

Since these points are not colinear, the image is indeed a circle.

So, if $z$ is the center, we have: $\mid z-1\mid=\mid z-\frac57\mid=\mid z-\frac{18+i}{25}\mid=r$.

This leads via a little algebra to $z=\frac67$. Thus $r=\frac17$.


Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.

In the present case, $w=f(z)$ means that $$w=\frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=\frac{2-3w}{4w-3}$$ Thus, the image of the circle has equation $$\left|\frac{2-3w}{4w-3}\right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+\bar w)+9|w|^2=16|w|^2-12(w+\bar w)+9$$ that is, $$7|w|^2-6(w+\bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=\frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.

Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...

  • 3
    $\begingroup$ +1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-\frac67(w+\bar{w})+\frac57=0$. Now "complete the square" to get $|w|^2-\frac67(w+\bar{w})+\frac{36}{49}=\frac{36}{49}-\frac57=\frac1{49}$. That's $\left(w-\frac67\right)\left(\bar{w}-\frac67\right)=\frac1{49}$. Now take the square root: $|w-\frac67|=\frac17$. $\endgroup$ – alex.jordan Jan 11 at 6:34
  • $\begingroup$ @alex.jordan Indeed. Well done and thanks. $\endgroup$ – Did Jan 11 at 9:02

Let see where this transformation takes $1,-1$ and $i$:

\begin{eqnarray} 1&\longmapsto &{5\over 7}\\-1&\longmapsto &1\\i&\longmapsto &{18+i\over 25}\\ \end{eqnarray}

Now calculate the center and radius of a triangle on $\alpha ={5\over 7}$, $\beta =1$ and $\gamma ={18+i\over 25}$.

Since this triangle is right at $\gamma$ we see that midpoint of segment $\alpha \beta$, that is $\sigma = {6\over 7}$ is a center of new circle with $r = {1\over 7}$.


Using Inversive Geometry

For a given LFT $\frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle $$ k\pm\frac{k+d/c}{|k+d/c|}r\tag1 $$ get mapped by the LFT to antipodal points of the image circle.

This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.

If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $\frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $\mathbb{C}$.

If one of the points computed in $(1)$ equals $-\frac dc$ (that is, that point is mapped to $\infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.

Given a pair of antipodal points on a circle, $\{p_1,p_2\}$, the radius, r, and center, k, of that circle are given by $$ r=\frac{|p_1-p_2|}2\qquad k=\frac{p_1+p_2}2\tag2 $$


In this case, we have $\frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle $$ 0\pm\frac{0+3/4}{|0+3/4|}\cdot1=\{-1,1\}\tag3 $$ get mapped by the LFT to the antipodal points of the image circle $$ \left\{1,\frac57\right\}\tag4 $$ then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are $$ \bbox[5px,border:2px solid #C0A000]{r=\frac17\qquad k=\frac67}\tag5 $$


The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.


Here's another solution I was able to find. Notice the matrix factorization:

$$ \left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right] = \left[ \begin{array}{cc} \frac{1}{5} & 0\\ 0 & 5 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array} \right] \times \left[ \begin{array}{cr} \frac{3}{5}& -\frac{4}{5}\\ \frac{4}{5}& \frac{3}{5} \end{array} \right] = A \times B \times C $$ The geometry behind this is that we have a Möbius transformation that factors into three parts: $$ \text{Möbius} = rotation \times translation \times dilation $$

Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = \pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:

$$ (-1,1) \stackrel{C}{\to} (7, - \frac{1}{7}) \stackrel{B}{\to} (25,\frac{125}{7})\stackrel{A}{\to} (1, \frac{5}{7}) $$ This corresponds to a circle centered at $z = \frac{6}{7}$ with radius $\frac{1}{7}$.

One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = \frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, \vec{u}) = (i, (1,0)) \in T_1(\mathbb{H}) $.

A Möbius transformation on $\mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(\mathbb{H})$ like this: $$ \left[ z \mapsto \frac{az+b}{cz+d} \right] \to \left[ (z, \vec{u}) \mapsto \left( \frac{az+b}{cz+d} , \frac{\vec{u}}{(cz+d)^2} \right) \right] $$ Let's see what happens when I try the previous example here: $$ \big(i, (1,0)\big) \mapsto \left( \frac{3i+2}{4i+3}, \frac{(1,0)}{(4i+3)^2}\right) = \left( \frac{18+i}{25} , \frac{1}{25}(24,-7) \right) $$ The factor of $\frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $\vec{u}$ would be tangent to a semi-circle with radius in the direction $\vec{u}_\perp$ passing through the point $f(z)=(\frac{18}{25}, \frac{1}{25})$. Therefore the center would be: $$ (\frac{18}{25}, \frac{1}{25}) + \frac{1}{7 \times 25}(24,-7) = (\frac{1}{7},0) $$ agreeing with the previous answer.


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